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Let $p(x) = a_0 + a_1 x^2 + a_2 x^4 + ... + a_n x^{2n}$ be a polynomial in a real variable $x$ with $0 < a_0 < a_1 < \ldots < a_n$. Prove that the function $p(x)$ has only one minimum.

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closed as off-topic by Davide Giraudo, William, Michael Albanese, Lost1, azimut Jul 13 '14 at 12:43

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2 Answers 2

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Outline: The fact that only even powers occur means that the curve $y=p(x)$ is symmetrical about the $y$-axis.

The fact that the $a_i$ are increasing is irrelevant. Since they are positive, if $|x|\lt |y|$ we have $\sum a_i x^{2i}\gt \sum a_i y^{2i}$.

Thus $p(x)$ has a local (and global) minimum only at $x=0$.

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Sir, I didn't really understand the part about symmetry about the y axis. –  user34304 Jul 13 '14 at 5:02
Since we are only using even powers, $p(-x)=p(x)$. I am taking a fairly geometric point of view, we could show there is a minimum only at $x=0$ with just algebra. –  André Nicolas Jul 13 '14 at 5:05

$\frac{dp(x)}{dx}|_{x=0}=\sum( \frac{d (a_i x^{2i})}{dx}|_{x=0})=\sum 0=0$ so $p(x)$ has a maximum or a minimum at $x=0$


$\frac{d^2 p(x)}{dx^2}|_{x=0}=\sum( \frac{d^2 (a_i x^{2i})}{dx^2}|_{x=0})>0$ since each $\frac{d^2 (a_i x^{2i})}{dx^2}|_{x=0}>0$

so it is a minimum

To show that it is the ONLY minimum, observe that

$\frac{d (a_i x^{2i})}{dx}<0$ for $x<0$ and $\frac{d (a_i x^{2i})}{dx}>0$ for $x>0$

so $\frac{dp(x)}{dx}<0$ for $x<0$ and $\frac{dp(x)}{dx}>0$ for $x>0$, so there are no other.

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I wouldn't agree with first sentence. Consider p(x) = $x^3$ –  RiaD Jul 13 '14 at 9:29

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